MATH 332 - ADVANCED CALCULUS
- Calculus of variations
Question Bank - Set 2
Liberty University
Question 1
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Euler-Lagrange equation. The extremals of the functional
J[y] satisfy the Euler-Lagrange equation:
d
dx ∂F
∂y′−∂F
∂y = 0,
where F=y′2−y.
Step 2: Compute the partial derivatives. We have
∂F
∂y′= 2y′,d
dx ∂F
∂y′= 2y′′.
And ∂F
∂y =−1.
Step 3: Substitute into the Euler-Lagrange equation. Substituting the par-
tial derivatives back into the Euler-Lagrange equation, we get
2y′′ + 1 = 0.
Step 4: Solve the differential equation. Solving the differential equation, we
find
y′′ =−1
2.
Step 5: Integrate twice. Integrating twice with respect to x, we get
y′=−x
2+C1,
y=−x2
4+C1x+C2.
Step 6: Apply boundary conditions. Using the boundary conditions y(0) = 0
and y(1) = 1, we find
y(0) = 0 =⇒C2= 0,
y(1) = 1 =⇒ −1
4+C1= 1 =⇒C1=5
4.
Step 7: Find the extremal. Therefore, the extremal that minimizes the
functional is
y=−x2
4+5
4x.
Question 2
Question
Find the stationary points of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Compute the Euler-Lagrange equation. The Euler-Lagrange equation
for the functional J[y] is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f(x, y, y′) = y′2−y2.
Step 2: Find the partial derivatives. We have ∂f
∂y′= 2y′and ∂f
∂y =−2y.
Step 3: Apply the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation to obtain
d
dx(2y′)−(−2y)=0
2
which simplifies to
2y′′ + 2y= 0
Step 4: Solve the differential equation. The general solution to the differen-
tial equation 2y′′ + 2y= 0 is
y=Asin√2x+Bcos√2x
where Aand Bare constants to be determined.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find
B= 0
Asin√2= 1
Step 6: Solve for the constants. From B= 0, we have y=Asin√2x.
Substituting y(1) = 1 into this equation gives Asin√2= 1, so
A=1
sin√2
Step 7: Final answer. Therefore, the stationary points of the functional J[y]
are given by
y(x) = 1
sin√2sin√2x
Question 3
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Euler-Lagrange equation. Given a functional of the form
J[y] = Rb
aF(x, y, y′)dx, the extremals y(x) are solutions to the Euler-Lagrange
equation:
d
dx ∂F
∂y′−∂F
∂y = 0
Step 2: Identify F(x, y, y′) from the given functional. Here, F(x, y, y′) =
y′2−y2.
3
Step 3: Calculate the partial derivatives of Fwith respect to yand y′.
∂F
∂y =−2y
∂F
∂y′= 2y′
Step 4: Differentiate ∂F
∂y′with respect to x.
d
dx ∂F
∂y′=d
dx(2y′)=2y′′
Step 5: Write the Euler-Lagrange equation.
2y′′ + 2y= 0
Step 6: Solve the differential equation obtained in Step 5. The general
solution to this differential equation is y(x) = c1cos(x)+c2sin(x), where c1and
c2are constants.
Step 7: Apply the boundary conditions to find the particular solution. Using
y(0) = 0, we have c1= 0. Using y(1) = 1, we have c2sin(1) = 1, so c2=1
sin(1) .
Therefore, the extremal that minimizes the given functional is y(x) = sin(x)
sin(1) .
Question 4
Question
Find the function y(x) that minimizes the functional
J[y] = Z1
0
(y′2−2y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Write the Euler-Lagrange equation. Given a functional of the form
J[y] = Zb
a
F(x, y, y′)dx,
where y′=dy/dx, the Euler-Lagrange equation for minimizing J[y] subject to
the boundary conditions y(a) = αand y(b) = βis
d
dx ∂F
∂y′=∂F
∂y .
4
In this case, F(x, y, y′) = y′2−2y, so
d
dx ∂(y′2−2y)
∂y′=∂(y′2−2y)
∂y .
Step 2: Compute the Euler-Lagrange equation. We have
d
dx (2y′) = −2,
which simplifies to
2y′′ =−2.
This gives us the second-order differential equation y′′ =−1.
Step 3: Solve the differential equation with the given boundary conditions.
Solving y′′ =−1 gives y(x) = −1
2x2+3
2x.
Applying the boundary conditions y(0) = 0 and y(1) = 1 gives the specific
solution y(x) = x.
Step 4: Check if this is the minimum. To verify if y(x) = xminimizes J[y],
we can use the second variation test or apply the boundary conditions to ensure
it is a minimum.
Therefore, the function y(x) = xminimizes the functional J[y].
Question 5
Question
Consider the functional J[y] = R1
0(y′2−2y)dx where y(0) = 0, y(1) = 1, and y
is twice differentiable. Find the function y(x) that minimizes J[y].
Solution
Step 1: Define the Euler-Lagrange equation for the given functional: The Euler-
Lagrange equation for this functional is given by d
dx ∂f
∂y′−∂f
∂y = 0, where
f(y, y′, x) = y′2−2y.
Step 2: Compute the partial derivatives of f:∂f
∂y′= 2y′and ∂f
∂y =−2.
Step 3: Substitute into the Euler-Lagrange equation: d
dx (2y′)−(−2) = 0.
This simplifies to 2y′′ + 2 = 0.
Step 4: Solve the differential equation: Divide by 2 to get y′′ + 1 = 0. The
general solution to this differential equation is y(x) = Ax +B, where Aand B
are constants.
Step 5: Apply the boundary conditions: Using y(0) = 0, we get 0 = B.
Therefore, B= 0.
Step 6: Apply the other boundary condition: Using y(1) = 1, we have
y(1) = A(1) + 0 = 1. This implies that A= 1.
Therefore, the function that minimizes the given functional is y(x) = x.
5
Question 6
Question
Find the extremals of the functional J[y] = R1
0(y′2−y2)dx subject to the
boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Let’s denote the integrand as L(y, y′, x) = y′2−y2.
Step 2: The Euler-Lagrange equation is given by d
dx ∂L
∂y′−∂L
∂y = 0.
Step 3: Compute ∂L
∂y′= 2y′and ∂L
∂y =−2y.
Step 4: Therefore, the Euler-Lagrange equation becomes d
dx (2y′)+2y= 0.
Step 5: Simplifying, we get 2y′′ + 2y= 0.
Step 6: The general solution to this differential equation is y(x) = c1sin(x)+
c2cos(x), where c1and c2are constants to be determined.
Step 7: Applying the boundary conditions y(0) = 0 and y(1) = 1, we find
c2= 0 and c1=1
sin(1) .
Step 8: Therefore, the extremal function that minimizes J[y] subject to the
given boundary conditions is y(x) = sin(x)
sin(1) .
Question 7
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Define the Euler-Lagrange equation. The extremals of the functional
J[y] satisfy the Euler-Lagrange equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−y2.
Step 2: Compute the partial derivatives. Let’s compute the partial deriva-
tives of fwith respect to yand y′:
∂f
∂y =−2yand ∂f
∂y′= 2y′
6
Step 3: Apply the Euler-Lagrange equation. Substitute these derivatives
into the Euler-Lagrange equation to obtain:
d
dx(2y′)+2y= 0
Simplify this differential equation to get:
2y′′ + 2y= 0
Step 4: Solve the differential equation. The general solution to this dif-
ferential equation is given by y(x) = c1cos(x) + c2sin(x), where c1and c2are
arbitrary constants.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 1 and y(1) = 2, we find:
y(0) = c1= 1
y(1) = c1cos(1) + c2sin(1) = 2
Solving these equations simultaneously, we get c2≈0.68294197.
Step 6: Final solution. Therefore, the extremal of the functional J[y] that
satisfies the boundary conditions is:
y(x) = cos(x)+0.68294197 sin(x)
Question 8
Question
Find the extremal for the functional
J[y] = Z1
0
(y′2+yy′)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Euler-Lagrange Equation The Euler-Lagrange equation for the
functional J[y] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
For F(x, y, y′) = y′2+yy′, we have
∂F
∂y =y′
∂F
∂y′= 2y′+y
7
Therefore, the Euler-Lagrange equation becomes
d
dx(2y′+y)−y′= 0
Step 2: Solve the Differential Equation Expanding and simplifying the
above equation, we get
2y′′ +y′−y′= 0
2y′′ = 0
y′′ = 0
Integrating twice, we find the general solution to be y(x) = Ax +B.
Step 3: Apply Boundary Conditions Using the boundary conditions
y(0) = 0 and y(1) = 1:
y(0) = A·0 + B= 0 =⇒B= 0
y(1) = A·1 + 0 = 1 =⇒A= 1
Step 4: Final Solution Thus, the extremal for the functional J[y] subject
to the given boundary conditions is y(x) = x.
Question 9
Question
Let J[y] = R1
0(y′2−y2)dx where y(0) = 0 and y(1) = 1. Find the function y(x)
that minimizes J[y].
Solution
Step 1: Compute the Euler-Lagrange equation.
The Euler-Lagrange equation for the functional J[y] = R1
0(y′2−y2)dx is
given by:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′) = y′2−y2.
Step 2: Compute the partial derivatives of fwith respect to yand y′.
∂f
∂y =−2yand ∂f
∂y′= 2y′
Step 3: Compute the Euler-Lagrange equation. Substituting the partial
derivatives into the Euler-Lagrange equation gives:
d
dx(2y′)−(−2y) = 0 =⇒d2y
dx2+ 2y= 0
8
Step 4: Solve the differential equation d2y
dx2+ 2y= 0.
The general solution to this differential equation is y(x) = c1cos√2x+
c2sin√2x.
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1.
Using y(0) = 0, we have c1= 0.
Using y(1) = 1, we have c2sin√2= 1. Hence, c2=1
sin(√2).
Therefore, the function that minimizes J[y] is y(x) = sin(√2x)
sin(√2).
Question 10
Question
Find the extremals for the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
To find the extremals of the functional, we will solve the Euler-Lagrange equa-
tion d
dx ∂f
∂y′−∂f
∂y = 0
where
f(x, y, y′) = y′2−y2
Step 1: Compute the partial derivatives of f. We have:
∂f
∂y′= 2y′
∂f
∂y =−2y
Step 2: Compute d
dx ∂f
∂y′.
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
Step 3: Set up the Euler-Lagrange equation. The Euler-Lagrange equation
is: d
dx ∂f
∂y′−∂f
∂y = 0
9
Substitute the derivatives of finto the equation:
2y′′ + 2y= 0
Step 4: Solve the differential equation. The general solution to 2y′′ +2y= 0
is given by
y(x) = c1cos(x) + c2sin(x)
Applying the boundary conditions y(0) = 0 and y(1) = 1 gives the specific
solution:
y(x) = sin(πx)
Question 11
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Lagrangian function. The Lagrangian function for the given
functional J[y] is given by
L(x, y, y′) = y′2−y2
Step 2: Apply the Euler-Lagrange equation. The Euler-Lagrange equation
for a functional of the form J[y] = Rb
aL(x, y, y′)dx is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: Compute the partial derivatives. Compute the partial derivatives of
L(x, y, y′):
∂L
∂y =−2y
∂L
∂y′= 2y′
d
dx ∂L
∂y′= 2y′′
Step 4: Apply the Euler-Lagrange equation. The Euler-Lagrange equation
becomes
2y′′ + 2y= 0
10
Step 5: Solve the differential equation. The general solution to the differen-
tial equation 2y′′ + 2y= 0 is
y(x) = c1cos(x) + c2sin(x)
Step 6: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find that
c1= 0
c2= 1
Step 7: Determine the extremal. Therefore, the extremal that minimizes the
functional J[y] is given by
y(x) = sin(x)
Question 12
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions
y(0) = 0
and
y(1) = 1
.
Solution
To find the extremals of the functional, we will use the Euler-Lagrange equation
given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 1: Compute ∂f
∂y′and ∂f
∂y .
∂f
∂y′= 2y′
∂f
∂y =−2y
11
Step 2: Plug ∂f
∂y′and ∂f
∂y into the Euler-Lagrange equation.
d
dx(2y′)+2y= 0
Step 3: Simplify the Euler-Lagrange equation.
2y′′ + 2y= 0
Step 4: Solve the differential equation 2y′′ + 2y= 0. The general solution
is
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1. From
y(0) = 0, we have:
0 = c1cos(0) + c2sin(0) = c1
Thus, y(x) = c2sin(x).
From y(1) = 1, we have:
1 = c2sin(1)
c2=1
sin(1)
Therefore, the extremal of the functional is
y(x) = 1
sin(1) sin(x)
.
Question 13
Question
Consider the functional
J[y] = Z1
0
(y′2+y2)dx
where the boundary conditions are y(0) = 0 and y(1) = 0. Find the function
y(x) that minimizes J[y].
Solution
Step 1: Define the Lagrangian for the functional J[y]. The Lagrangian for the
given functional can be defined as
L(x, y, y′) = y′2+y2+λ1y+λ2(y−1)
where λ1and λ2are the Lagrange multipliers corresponding to the boundary
constraints y(0) = 0 and y(1) = 0 respectively.
12
Step 2: Find the Euler-Lagrange equation. The Euler-Lagrange equation for
this problem is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Plugging in the expression for L, we get
d
dx(2y′)−2y−λ1−λ2= 0
Step 3: Solve the Euler-Lagrange equation. Solving the differential equation,
we obtain
2y′′ −2y=λ1+λ2
The general solution to this differential equation is
y(x) = c1sin(x) + c2cos(x) + 1
2(λ1+λ2)
Step 4: Apply the boundary conditions. From the boundary conditions
y(0) = 0 and y(1) = 0, we get
c2+1
2(λ1+λ2) = 0
c1sin(1) + c2cos(1) + 1
2(λ1+λ2)=0
Solving these equations simultaneously, we find the function that minimizes J[y]
is
y(x) = 2
sin(1) sin(x)−1
Question 14
Question
Find the function y(x) that minimizes the functional
J[y] = Z2
0
[(y′)2+y2]dx
subject to the boundary conditions y(0) = 0 and y(2) = 1.
Solution
Step 1: Define the Lagrangian function.
The Lagrangian function for this problem is given by
L(x, y, y′)=(y′)2+y2+λ1(0)y+λ2(2)(y−1)
13
where λ1and λ2are the Lagrange multipliers corresponding to the boundary
conditions.
Step 2: Find the Euler-Lagrange equation.
The Euler-Lagrange equation is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Plugging in values and simplifying, we get
d
dx(2y′)−2y= 0
Step 3: Solve the Euler-Lagrange equation.
Solving the differential equation, we get
y′′ =y
The general solution to this differential equation is
y(x) = c1ex+c2e−x
Step 4: Apply boundary conditions.
Using the boundary conditions y(0) = 0 and y(2) = 1, we find that c1=1
e2−e−2
and c2=−1
e2−e−2.
Therefore, the function that minimizes the given functional is
y(x) = ex−e−x
e2−e−2
Question 15
Question
Find the extremal of the functional
J[y] = Z1
0
(y′2−2y2)dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
the functional J[y] = Rb
aL(x, y, y′)dx is given by
d
dx ∂L
∂y′−∂L
∂y = 0
14
In this case, our Lagrangian function is L(x, y, y′) = y′2−2y2, so we have
d
dx (2y′)−(−4y) = 0
which simplifies to d
dx (2y′)+4y= 0
Step 2: Solve the differential equation. Rewriting the equation in a more
familiar form, we have
2y′′ + 4y= 0
The characteristic equation is 2r2+ 4 = 0, which gives us r=±i.
Therefore, the general solution is
y(x) = c1cos(2x) + c2sin(2x)
Step 3: Apply boundary conditions.
From the first boundary condition y(0) = 1, we have c1= 1. From the
second boundary condition y(1) = 2, we have c1cos(2) + c2sin(2) = 2.
Substitute c1= 1 into the second boundary condition to get
cos(2) + c2sin(2) = 2
Solving for c2, we get
c2=2−cos(2)
sin(2)
Step 4: Final solution.
Therefore, the extremal of the functional is
y(x) = cos(2x) + 2−cos(2)
sin(2) sin(2x)
Question 16
Question
Find the extremals of the functional
J[y] = Z1
0
(y2+y′2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
15
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
the functional J[y] = Rb
aL(x, y, y′)dx is given by
d
dx
∂L
∂y′−∂L
∂y = 0.
Step 2: Given the functional J[y] = R1
0(y2+y′2)dx, we have L(x, y, y′) =
y2+y′2. The Euler-Lagrange equation becomes
d
dx
∂(y2+y′2)
∂y′−∂(y2+y′2)
∂y = 0.
Step 3: Compute the partial derivatives. We have
∂(y2+y′2)
∂y′= 2y′,
∂(y2+y′2)
∂y = 2y.
Substituting these into the Euler-Lagrange equation gives
d
dx(2y′)−2y= 0.
Step 4: Simplify the Euler-Lagrange equation. This simplifies to
2y′′ −2y= 0.
Step 5: Solve the differential equation. The general solution to 2y′′ −2y= 0
is y(x) = c1cos√2x+c2sin√2x.
Step 6: Apply the boundary conditions. Using y(0) = 0, we find c1= 0.
Using y(1) = 1, we find c2=1
sin(√2).
Step 7: The extremal for the functional is
y(x) = sin√2x
sin√2.
Question 17
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
16
Solution
Step 1: Compute the Euler-Lagrange equation. The Euler-Lagrange equation
for a functional of the form J[y] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
In this case, F(x, y, y′) = y′2−y2, so
∂F
∂y =−2yand ∂F
∂y′= 2y′
d
dx ∂F
∂y′=d
dx(2y′)=2y′′
Therefore, the Euler-Lagrange equation is
2y′′ + 2y= 0
Step 2: Solve the differential equation. The general solution to the differen-
tial equation 2y′′ + 2y= 0 is
y(x) = c1cos(x) + c2sin(x)
Step 3: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we have
y(0) = c1= 0
y(1) = c2sin(1) = 1
c2=1
sin(1)
Therefore, the extremal that minimizes the functional is
y(x) = 1
sin(1) sin(x)
Question 18
Question
Find the extremals for the functional
J[y] = Z1
0
(y′2+y′′)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
17
Solution
Step 1: Compute the Euler-Lagrange equation. Given the functional
J[y] = Z1
0
(y′2+y′′)dx
We have the Euler-Lagrange equation
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2+y′′.
Step 2: Find the partial derivatives and substitute into the Euler-Lagrange
equation. We have ∂f
∂y′=∂
∂y′(y′2+y′′)=2y′
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
∂f
∂y =∂
∂y (y′2+y′′)=0
Substitute these derivatives into the Euler-Lagrange equation:
2y′′ −0=0
y′′ = 0
Step 3: Solve the differential equation y′′ = 0. The general solution to y′′ = 0
is y=Ax +B, where Aand Bare constants.
Step 4: Apply the boundary conditions. Since y(0) = 0 and y(1) = 1, we
have the following system of equations:
B= 0
A+B= 1
From the first equation, we have B= 0, and substituting into the second
equation gives A= 1. Therefore, the extremal that minimizes the functional is
y=x.
Question 19
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
18
Solution
Step 1: Compute the Euler-Lagrange equation. This equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y.
Step 2: Compute the partial derivatives:
∂f
∂y′=∂(y′2−y)
∂y′= 2y′
∂f
∂y =∂(y′2−y)
∂y =−1
Step 3: Plug the derivatives into the Euler-Lagrange equation:
d
dx (2y′) + 1 = 0
Step 4: Simplify the equation:
2y′′ + 1 = 0
Step 5: Solve the differential equation:
2y′′ =−1
y′′ =−1
2
Step 6: Integrate y′′ twice to find y:
y′=−x
2+A
y=−x2
4+Ax +B
Step 7: Apply the boundary conditions to solve for Aand B:
y(0) = 0 ⇒B= 0
y(1) = 1 ⇒ −1
4+A= 1 ⇒A=5
4
Therefore, the extremal for the functional J[y] that satisfies the boundary
conditions is
y=−x2
4+5
4x
19
Question 20
Question
Let J[y] = R1
0y(x)−y′(x)2dx where y(0) = 0 and y(1) = 1. Find the function
y(x) that minimizes J[y].
Solution
Step 1: Write down the Euler-Lagrange equation for the given functional J[y]:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y−y′(x)2.
Step 2: Compute the partial derivatives:
∂f
∂y = 1 and ∂f
∂y′=−2y′
Step 3: Apply the Euler-Lagrange equation:
d
dx(−2y′)−1=0
Simplify the equation to get:
2y′′ −1=0
Step 4: Solve the differential equation 2y′′ −1 = 0 subject to the boundary
conditions y(0) = 0 and y(1) = 1.
Step 5: Integrate twice to find the general solution:
y′′ =1
2=⇒y′=1
2x+C1
y=1
4x2+C1x+C2
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1 to find C1
and C2:
0 = C2and 1 = 1
4+C1
Thus, C1=3
4.
Step 7: Substitute C1back into yto get the specific function that minimizes
J[y]:
y=1
4x2+3
4x
Therefore, the function y(x) = 1
4x2+3
4xminimizes the functional J[y].
20
Question 21
Question
Find the extremals for the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
To find the extremals for the functional J[y], we will apply the Euler-Lagrange
equation.
Step 1: Formulating the Euler-Lagrange equation Let F(x, y, y′) =
y′2−y2. The Euler-Lagrange equation is given by
d
dx ∂F
∂y′−∂F
∂y = 0
Step 2: Computing the partial derivatives Compute the partial deriva-
tives of F:∂F
∂y′= 2y′,∂F
∂y =−2y
Step 3: Applying the Euler-Lagrange equation Substitute the partial
derivatives into the Euler-Lagrange equation:
d
dx(2y′)−(−2y)=0
d
dx(2y′)+2y= 0
Step 4: Solving the differential equation Integrate the above equation:
2y′+ 2y=C
y′+y=C1
where C1is an arbitrary constant.
Step 5: Determining the extremals Applying the boundary conditions
y(0) = 0 and y(1) = 1:
y′(0) + y(0) = C1= 0
y′(1) + y(1) = C1= 1
Thus, the extremal that satisfies these boundary conditions is y(x) = e−x.
Therefore, the extremal for the functional J[y] is y(x) = e−x.
21
Question 22
Question
Consider the functional J(y) = R1
0(y′2−y2)dx where the function ysatisfies the
boundary conditions y(0) = 0 and y(1) = 1. Find the function ythat minimizes
J(y).
Solution
Step 1: Calculate the Euler-Lagrange equation by differentiating the integrand
with respect to yand y′.
d
dx ∂f
∂y′−∂f
∂y = 0
In this case, the Lagrangian f=y′2−y2, so ∂f
∂y =−2yand ∂f
∂y′= 2y′. Then,
d
dx(2y′)−(−2y)=0
2y′′ + 2y= 0
y′′ +y= 0
Step 2: Solve the differential equation y′′ +y= 0 subject to the boundary
conditions y(0) = 0 and y(1) = 1.
The general solution to the differential equation is of the form y(x) =
Asin x+Bcos x. Applying the boundary conditions: y(0) = 0 ⇒B= 0
y(1) = 1 ⇒Asin 1 = 1 ⇒A=1
sin 1
Therefore, the function ythat minimizes the functional J(y) is y(x) = sin x
sin 1 .
Question 23
Question
Let J[y] = R1
0(y′2−y2)dx be a functional defined on the set of twice continuously
differentiable functions y(x) satisfying y(0) = 0 and y(1) = 1. Find the function
y(x) that minimizes J[y].
Solution
Step 1: Let L(y, y′, x) = y′2−y2be the integrand of the functional J[y].
Step 2: The Euler-Lagrange equation for minimizing J[y] is given by
d
dx ∂L
∂y′−∂L
∂y = 0
22
Step 3: Compute the partial derivatives:
∂L
∂y′= 2y′
∂L
∂y =−2y
Step 4: Apply the Euler-Lagrange equation:
d
dx (2y′)−(−2y) = 0
2y′′ + 2y= 0
Step 5: Solve the differential equation: The general solution to y′′ +y= 0
is of the form y(x) = Asin(x) + Bcos(x).
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1:
y(0) = B= 0
y(1) = Asin(1) = 1
A=1
sin(1)
Step 7: Therefore, the function y(x) = sin(x)
sin(1) minimizes the functional J[y]
subject to the given boundary conditions.
Question 24
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2+yy′+y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Calculate the Euler-Lagrange equation by differentiating the integrand
with respect to yand y′.
∂f
∂y −d
dx ∂f
∂y′= 0
where f=y′2+yy′+y2.
23
Step 2: Compute ∂f
∂y and ∂f
∂y′.
∂f
∂y = 2y+y′
∂f
∂y′= 2y′+y
Step 3: Differentiate ∂f
∂y′with respect to x.
d
dx ∂f
∂y′=d
dx(2y′+y)=2y′′ +y′
Step 4: Set up the Euler-Lagrange equation and simplify.
2y+y′−(2y′′ +y′) = 0
2y+y′−2y′′ −y′= 0
2y−2y′′ = 0
y−y′′ = 0
Step 5: Solve the differential equation y−y′′ = 0 with the boundary condi-
tions y(0) = 0 and y(1) = 1. The general solution to the differential equation is
y(x) = c1ex+c2e−x. Using the boundary conditions:
y(0) = c1+c2= 0
y(1) = c1e+c2e−1= 1
Step 6: Solve the system of equations to find the constants c1and c2.
c1=e
e−1, c2=−1
e−1
Thus, the extremal for the given functional is y(x) = e
e−1ex−1
e−1e−x.
Question 25
Question
Find the extremals of the functional
J[y] = Z1
0
(2y2+y′2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
24
Solution
Step 1: Define the Euler-Lagrange equation.
The Euler-Lagrange equation for a functional J[y] = Rb
aL(x, y, y′)dx is given
by: d
dx
∂L
∂y′=∂L
∂y
Step 2: Find the partial derivatives of L(x, y, y′).
In this case, L(x, y, y′)=2y2+y′2, so:
∂L
∂y = 4yand ∂L
∂y′= 2y′
Step 3: Calculate the derivative d
dx
∂L
∂y′.
Taking the derivative of ∂L
∂y′= 2y′with respect to x:
d
dx
∂L
∂y′=d
dx(2y′)=2y′′
Step 4: Apply the Euler-Lagrange equation.
By applying the Euler-Lagrange equation, we have:
2y′′ = 4y
This simplifies to the second-order differential equation y′′ −2y= 0.
Step 5: Solve the differential equation y′′ −2y= 0.
The characteristic equation for this differential equation is r2−2 = 0, which has
solutions r=±√2. Therefore, the general solution to the differential equation
is:
y(x) = c1e√2x+c2e−√2x
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1.
Using the boundary conditions y(0) = 0 and y(1) = 1, we find:
0 = c1+c2
1 = c1e√2+c2e−√2
Solving this system of equations, we get c1=e−√2
e−√2−e√2and c2=−e√2
e−√2−e√2.
Step 7: Determine the extremal function.
Substitute the values of c1and c2back into the general solution to obtain the
extremal function y(x).
Therefore, the extremals of the functional J[y] = R1
0(2y2+y′2)dx subject
to the boundary conditions y(0) = 0 and y(1) = 1 are the solutions to the
differential equation y′′ −2y= 0 with the extremal function derived from the
boundary conditions.
25
Question 26
Question
Find the extremals for the functional
J[y] = Z1
0
(2y−y′2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
To find the extremals for the given functional J[y], we will use the Euler-
Lagrange equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f= 2y−y′2.
Step 1: Find ∂f ∂y and ∂f
∂y′
∂f
∂y = 2
∂f
∂y′=−2y′
Step 2: Apply the Euler-Lagrange equation
d
dx(−2y′)−2=0
d
dx(−2y′)=2
−2y′′ = 2
Step 3: Solve the ODE Integrate both sides to solve for y(x):
−2y′= 2x+C1
y′=−x−C1/2
y=−1
2x2−C1x+C2
Step 4: Apply the boundary conditions Using the boundary conditions
y(0) = 0 and y(1) = 1:
y(0) = C2= 0
y(1) = −1
2+C1= 1
26
C1=3
2
Step 5: Final Solution The extremal for the given functional subject to
the boundary conditions is:
y(x) = −1
2x2+3
2x
Question 27
Question
Find the extremals of the functional
J[y] = Z1
0
(y′(x))2+y(x)2dx
subject to the boundary conditions y(0) = 0 and y(1) = 3.
Solution
Step 1: Define the Euler-Lagrange equation The Euler-Lagrange equation for
the given functional J[y] is
d
dx ∂f
∂y′−∂f
∂y = 0,
where f(y, y′, x)=(y′(x))2+y(x)2.
Step 2: Calculate the partial derivatives Let’s first find the partial derivatives
of fwith respect to yand y′:
∂f
∂y = 2y(x),∂f
∂y′= 2y′(x).
Step 3: Apply the Euler-Lagrange equation Substitute the partial derivatives
into the Euler-Lagrange equation:
d
dx(2y′(x)) −2y(x)=0
2y′′(x)−2y(x) = 0.
Step 4: Solve the differential equation The general solution to the differential
equation 2y′′(x)−2y(x) = 0 is
y(x) = c1cos(x) + c2sin(x).
Step 5: Apply the boundary conditions Using the boundary conditions
y(0) = 0 and y(1) = 3, we find:
(y(0) = c1= 0
y(1) = c1cos(1) + c2sin(1) = 3
27
Step 6: Solve for the constants From c1= 0, we have c2sin(1) = 3, so
c2=3
sin(1) . Thus, the extremal of the functional is
y(x) = 3
sin(1) sin(x).
Question 28
Question
Find the extremal for the functional J[y] = R1
0y′(x)2+y(x)2dx subject to the
constraint R1
0y(x)dx = 1.
Solution
Step 1: Define the Lagrangian. Let L=y′(x)2+y(x)2+λ(y(x)−1), where λ
is a Lagrange multiplier.
Step 2: Find the Euler-Lagrange equation. The Euler-Lagrange equation
for this problem is given by
d
dx ∂L
∂y′−∂L
∂y = 0.
This gives us
d
dx(2y′(x)) −2y(x) + λ= 0 and 2y(x)+2λ= 0.
Step 3: Solve for y(x) using the boundary condition. From the second
equation in Step 2, we have y(x) = −λ. Substituting this back into the first
equation and integrating, we find y(x) = sin 2πx.
Step 4: Use the constraint to find λ. Apply the constraint R1
0y(x)dx = 1:
Z1
0
sin 2πx dx =1
2πcos 2πx
1
0=1
π= 1.
Thus, λ=−1.
Step 5: Final extremal. Therefore, the extremal for the given functional
subject to the constraint is y(x) = −sin 2πx.
Question 29
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
28
Solution
Step 1: Define the Lagrangian function. We define the Lagrangian function as
L(x, y, y′) = y′2−y2
Step 2: Set up the Euler-Lagrange equation. The Euler-Lagrange equation
for the functional J[y] is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Applying this to our Lagrangian L(x, y, y′) = y′2−y2, we get
d
dx(2y′)+2y= 0
Step 3: Solve the Euler-Lagrange equation. Solving the above differential
equation gives us
2y′′ + 2y= 0
Step 4: Find the general solution to the differential equation. The general
solution to the differential equation 2y′′ + 2y= 0 is
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find
c1= 0
c2sin(1) = 1
c2=1
sin(1)
Step 6: Write down the extremal. Therefore, the extremal of the functional
J[y] that satisfies the given boundary conditions is
y(x) = 1
sin(1) sin(x)
Question 30
Question
Find the extremals for the functional
J[y] = Z2
1
(y′2−2y)dx
subject to the boundary conditions y(1) = 0 and y(2) = 1.
29
Solution
Step 1: Compute the Euler-Lagrange equation. The Euler-Lagrange equation
for the functional J[y] is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−2y.
Step 2: Find the partial derivatives. We have
∂f
∂y =−2,and ∂f
∂y′= 2y′
and d
dx ∂f
∂y′=d
dx(2y′)=2y′′
Step 3: Form the Euler-Lagrange equation. Substitute the partial derivatives
into the Euler-Lagrange equation to get
2y′′ + 2 = 0
Step 4: Solve the differential equation. The general solution to the differen-
tial equation is
y(x) = c1+c2x−x2
where c1and c2are constants to be determined.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(1) = 0 and y(2) = 1, we have
c1+c2−1=0
c1+ 2c2−4=1
Solving these equations simultaneously, we find c1=−2 and c2= 3.
Therefore, the extremal for the functional J[y] is
y(x) = −2+3x−x2
Question 31
Question
Find the extremal of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
30
Solution
Step 1: We define the Euler-Lagrange equation as
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 2: Calculate the partial derivatives:
∂f
∂y′= 2y′and ∂f
∂y =−2y
Step 3: Compute the derivative with respect to xin the Euler-Lagrange
equation:
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
Step 4: Substitute the derivatives back into the Euler-Lagrange equation:
2y′′ −(−2y)=0
2y′′ + 2y= 0
Step 5: Solve the differential equation 2y′′ + 2y= 0 for y(x) with the bound-
ary conditions y(0) = 0 and y(1) = 1.
Step 6: The general solution to the differential equation is y(x) = Asin(x) +
Bcos(x).
Step 7: Use the boundary conditions to find Aand B:
y(0) = 0 =⇒B= 0
y(1) = 1 =⇒Asin(1) = 1 =⇒A=1
sin(1)
Step 8: Therefore, the extremal of the functional J[y] subject to the given
boundary conditions is
y(x) = sin(x)
sin(1)
Question 32
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
31
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
the functional J[y] = Rb
aL(x, y, y′)dx is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Step 2: Calculate the partial derivatives. In this case, we have L(x, y, y′) =
y′2−y. Therefore, the partial derivatives are:
∂L
∂y =−1
∂L
∂y′= 2y′
d
dx ∂L
∂y′= 2y′′
Step 3: Set up the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation to get:
d
dx(2y′) + 1 = 0
Step 4: Solve the differential equation. The differential equation simplifies
to 2y′′ + 1 = 0. Solving this gives y(x) = 1
2x2+Ax +B.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find B= 0 and A=1
2. Therefore, the extremal is
y(x) = 1
2x2+1
2x.
Step 6: Check the second variation. To confirm that this extremal is a
minimum, we need to check the second variation. This involves computing the
second variation of the functional and verifying the sign of δ2J.
Question 33
Question
Evaluate the functional
J[y] = Z1
02y2+y′2dx
subject to the boundary conditions y(0) = 0 and y(1) = 0.
32
Solution
Step 1: Compute the Euler-Lagrange equation by differentiating the integrand
with respect to yand y′.
∂f
∂y −d
dx ∂f
∂y′= 0
where f= 2y2+y′2.
Step 2: Compute ∂f
∂y .
∂f
∂y = 4y
Step 3: Compute ∂f
∂y′.
∂f
∂y′= 2y′
Step 4: Compute d
dx ∂f
∂y′.
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
Step 5: Substitute the results back into the Euler-Lagrange equation.
4y−2y′′ = 0
Step 6: Rearrange the equation to get a differential equation.
2y′′ −4y= 0
Step 7: Solve the differential equation by assuming a solution of the form
y(x) = erx.
2r2erx −4erx = 0
Step 8: Simplify the equation and solve for r.
2r2−4 = 0 =⇒r=±√2
Step 9: The general solution is of the form y(x) = c1e√2x+c2e−√2x.
Step 10: Apply the boundary conditions y(0) = 0 and y(1) = 0 to find the
specific solution.
c1+c2= 0
c1e√2+c2e−√2= 0
Step 11: Solve the system of equations to find c1and c2.
c1=e−√2
e√2−e−√2=−1
2√2
c2=e√2
e√2−e−√2=1
2√2
Step 12: Therefore, the function that minimizes the functional J[y] is y(x) =
−1
2√2e√2x+1
2√2e−√2x.
33
Question 34
Question
Consider the functional J[y] = R1
0(y′(x))2+y(x)dx subject to the boundary
conditions y(0) = 1 and y(1) = 2. Find the function y(x) that minimizes J[y].
Solution
Step 1: Define the Lagrangian. The Lagrangian for the given functional J[y]
with the constraint of fixed boundary conditions can be defined as:
L(y, y′;λ)=(y′(x))2+y(x) + λ1(y(0) −1) + λ2(y(1) −2)
Step 2: Find the Euler-Lagrange equation. The Euler-Lagrange equation for
minimizing the functional J[y] is given by:
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: Solve the Euler-Lagrange equation. Applying the Euler-Lagrange
equation to the Lagrangian L, we get:
d
dx(2y′)−1=0
2y′′ −1=0
Step 4: Solve the differential equation. Solving the differential equation
2y′′ −1 = 0 with the boundary conditions y(0) = 1 and y(1) = 2, we find:
y(x) = 1
2x2+3
2x+ 1
Step 5: Verify the solution. To verify that this function minimizes the
functional J[y], we need to check if it satisfies the boundary conditions and
compare its value with other candidate functions.
Question 35
Question
Find the extremals for the functional
J[y] = Z2
1
(y′2−2y)dx
subject to the boundary conditions y(1) = 1 and y(2) = 2.
34
Solution
Step 1: Compute the Euler-Lagrange equation. The Euler-Lagrange equation
for the functional J[y] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
In this case, F(x, y, y′) = y′2−2y. Thus,
∂F
∂y =−2 and ∂F
∂y′= 2y′
Therefore, the Euler-Lagrange equation becomes
d
dx(2y′) + 2 = 0
2y′′ + 2 = 0
Step 2: Solve the differential equation. Solving the differential equation
2y′′ + 2 = 0 gives us
y′′ =−1
Integrating once gives
y′=−x+c1
Integrating again gives
y=−1
2x2+c1x+c2
Step 3: Apply the boundary conditions. Using the first boundary condition
y(1) = 1,
1 = −1
2(1)2+c1(1) + c2
1 = −1
2+c1+c2
Using the second boundary condition y(2) = 2,
2 = −1
2(2)2+c1(2) + c2
2 = −2+2c1+c2
Solving the system of equations, we find c1=3
2and c2= 1.
Therefore, the extremal function is
y=−1
2x2+3
2x+ 1
35
Question 6
Question
Find the extremals of the functional J[y] = R1
0(y′2−y2)dx subject to the
boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Let’s denote the integrand as L(y, y′, x) = y′2−y2.
Step 2: The Euler-Lagrange equation is given by d
dx ∂L
∂y′−∂L
∂y = 0.
Step 3: Compute ∂L
∂y′= 2y′and ∂L
∂y =−2y.
Step 4: Therefore, the Euler-Lagrange equation becomes d
dx (2y′)+2y= 0.
Step 5: Simplifying, we get 2y′′ + 2y= 0.
Step 6: The general solution to this differential equation is y(x) = c1sin(x)+
c2cos(x), where c1and c2are constants to be determined.
Step 7: Applying the boundary conditions y(0) = 0 and y(1) = 1, we find
c2= 0 and c1=1
sin(1) .
Step 8: Therefore, the extremal function that minimizes J[y] subject to the
given boundary conditions is y(x) = sin(x)
sin(1) .
Question 7
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Define the Euler-Lagrange equation. The extremals of the functional
J[y] satisfy the Euler-Lagrange equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−y2.
Step 2: Compute the partial derivatives. Let’s compute the partial deriva-
tives of fwith respect to yand y′:
∂f
∂y =−2yand ∂f
∂y′= 2y′
6
Step 3: Apply the Euler-Lagrange equation. Substitute these derivatives
into the Euler-Lagrange equation to obtain:
d
dx(2y′)+2y= 0
Simplify this differential equation to get:
2y′′ + 2y= 0
Step 4: Solve the differential equation. The general solution to this dif-
ferential equation is given by y(x) = c1cos(x) + c2sin(x), where c1and c2are
arbitrary constants.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 1 and y(1) = 2, we find:
y(0) = c1= 1
y(1) = c1cos(1) + c2sin(1) = 2
Solving these equations simultaneously, we get c2≈0.68294197.
Step 6: Final solution. Therefore, the extremal of the functional J[y] that
satisfies the boundary conditions is:
y(x) = cos(x)+0.68294197 sin(x)
Question 8
Question
Find the extremal for the functional
J[y] = Z1
0
(y′2+yy′)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Euler-Lagrange Equation The Euler-Lagrange equation for the
functional J[y] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
For F(x, y, y′) = y′2+yy′, we have
∂F
∂y =y′
∂F
∂y′= 2y′+y
7
Therefore, the Euler-Lagrange equation becomes
d
dx(2y′+y)−y′= 0
Step 2: Solve the Differential Equation Expanding and simplifying the
above equation, we get
2y′′ +y′−y′= 0
2y′′ = 0
y′′ = 0
Integrating twice, we find the general solution to be y(x) = Ax +B.
Step 3: Apply Boundary Conditions Using the boundary conditions
y(0) = 0 and y(1) = 1:
y(0) = A·0 + B= 0 =⇒B= 0
y(1) = A·1 + 0 = 1 =⇒A= 1
Step 4: Final Solution Thus, the extremal for the functional J[y] subject
to the given boundary conditions is y(x) = x.
Question 9
Question
Let J[y] = R1
0(y′2−y2)dx where y(0) = 0 and y(1) = 1. Find the function y(x)
that minimizes J[y].
Solution
Step 1: Compute the Euler-Lagrange equation.
The Euler-Lagrange equation for the functional J[y] = R1
0(y′2−y2)dx is
given by:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′) = y′2−y2.
Step 2: Compute the partial derivatives of fwith respect to yand y′.
∂f
∂y =−2yand ∂f
∂y′= 2y′
Step 3: Compute the Euler-Lagrange equation. Substituting the partial
derivatives into the Euler-Lagrange equation gives:
d
dx(2y′)−(−2y) = 0 =⇒d2y
dx2+ 2y= 0
8
Step 4: Solve the differential equation d2y
dx2+ 2y= 0.
The general solution to this differential equation is y(x) = c1cos√2x+
c2sin√2x.
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1.
Using y(0) = 0, we have c1= 0.
Using y(1) = 1, we have c2sin√2= 1. Hence, c2=1
sin(√2).
Therefore, the function that minimizes J[y] is y(x) = sin(√2x)
sin(√2).
Question 10
Question
Find the extremals for the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
To find the extremals of the functional, we will solve the Euler-Lagrange equa-
tion d
dx ∂f
∂y′−∂f
∂y = 0
where
f(x, y, y′) = y′2−y2
Step 1: Compute the partial derivatives of f. We have:
∂f
∂y′= 2y′
∂f
∂y =−2y
Step 2: Compute d
dx ∂f
∂y′.
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
Step 3: Set up the Euler-Lagrange equation. The Euler-Lagrange equation
is: d
dx ∂f
∂y′−∂f
∂y = 0
9
Substitute the derivatives of finto the equation:
2y′′ + 2y= 0
Step 4: Solve the differential equation. The general solution to 2y′′ +2y= 0
is given by
y(x) = c1cos(x) + c2sin(x)
Applying the boundary conditions y(0) = 0 and y(1) = 1 gives the specific
solution:
y(x) = sin(πx)
Question 11
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Define the Lagrangian function. The Lagrangian function for the given
functional J[y] is given by
L(x, y, y′) = y′2−y2
Step 2: Apply the Euler-Lagrange equation. The Euler-Lagrange equation
for a functional of the form J[y] = Rb
aL(x, y, y′)dx is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: Compute the partial derivatives. Compute the partial derivatives of
L(x, y, y′):
∂L
∂y =−2y
∂L
∂y′= 2y′
d
dx ∂L
∂y′= 2y′′
Step 4: Apply the Euler-Lagrange equation. The Euler-Lagrange equation
becomes
2y′′ + 2y= 0
10
Step 5: Solve the differential equation. The general solution to the differen-
tial equation 2y′′ + 2y= 0 is
y(x) = c1cos(x) + c2sin(x)
Step 6: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find that
c1= 0
c2= 1
Step 7: Determine the extremal. Therefore, the extremal that minimizes the
functional J[y] is given by
y(x) = sin(x)
Question 12
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions
y(0) = 0
and
y(1) = 1
.
Solution
To find the extremals of the functional, we will use the Euler-Lagrange equation
given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 1: Compute ∂f
∂y′and ∂f
∂y .
∂f
∂y′= 2y′
∂f
∂y =−2y
11
Step 2: Plug ∂f
∂y′and ∂f
∂y into the Euler-Lagrange equation.
d
dx(2y′)+2y= 0
Step 3: Simplify the Euler-Lagrange equation.
2y′′ + 2y= 0
Step 4: Solve the differential equation 2y′′ + 2y= 0. The general solution
is
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions y(0) = 0 and y(1) = 1. From
y(0) = 0, we have:
0 = c1cos(0) + c2sin(0) = c1
Thus, y(x) = c2sin(x).
From y(1) = 1, we have:
1 = c2sin(1)
c2=1
sin(1)
Therefore, the extremal of the functional is
y(x) = 1
sin(1) sin(x)
.
Question 13
Question
Consider the functional
J[y] = Z1
0
(y′2+y2)dx
where the boundary conditions are y(0) = 0 and y(1) = 0. Find the function
y(x) that minimizes J[y].
Solution
Step 1: Define the Lagrangian for the functional J[y]. The Lagrangian for the
given functional can be defined as
L(x, y, y′) = y′2+y2+λ1y+λ2(y−1)
where λ1and λ2are the Lagrange multipliers corresponding to the boundary
constraints y(0) = 0 and y(1) = 0 respectively.
12
Step 2: Find the Euler-Lagrange equation. The Euler-Lagrange equation for
this problem is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Plugging in the expression for L, we get
d
dx(2y′)−2y−λ1−λ2= 0
Step 3: Solve the Euler-Lagrange equation. Solving the differential equation,
we obtain
2y′′ −2y=λ1+λ2
The general solution to this differential equation is
y(x) = c1sin(x) + c2cos(x) + 1
2(λ1+λ2)
Step 4: Apply the boundary conditions. From the boundary conditions
y(0) = 0 and y(1) = 0, we get
c2+1
2(λ1+λ2) = 0
c1sin(1) + c2cos(1) + 1
2(λ1+λ2)=0
Solving these equations simultaneously, we find the function that minimizes J[y]
is
y(x) = 2
sin(1) sin(x)−1
Question 14
Question
Find the function y(x) that minimizes the functional
J[y] = Z2
0
[(y′)2+y2]dx
subject to the boundary conditions y(0) = 0 and y(2) = 1.
Solution
Step 1: Define the Lagrangian function.
The Lagrangian function for this problem is given by
L(x, y, y′)=(y′)2+y2+λ1(0)y+λ2(2)(y−1)
13
where λ1and λ2are the Lagrange multipliers corresponding to the boundary
conditions.
Step 2: Find the Euler-Lagrange equation.
The Euler-Lagrange equation is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Plugging in values and simplifying, we get
d
dx(2y′)−2y= 0
Step 3: Solve the Euler-Lagrange equation.
Solving the differential equation, we get
y′′ =y
The general solution to this differential equation is
y(x) = c1ex+c2e−x
Step 4: Apply boundary conditions.
Using the boundary conditions y(0) = 0 and y(2) = 1, we find that c1=1
e2−e−2
and c2=−1
e2−e−2.
Therefore, the function that minimizes the given functional is
y(x) = ex−e−x
e2−e−2
Question 15
Question
Find the extremal of the functional
J[y] = Z1
0
(y′2−2y2)dx
subject to the boundary conditions y(0) = 1 and y(1) = 2.
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
the functional J[y] = Rb
aL(x, y, y′)dx is given by
d
dx ∂L
∂y′−∂L
∂y = 0
14
In this case, our Lagrangian function is L(x, y, y′) = y′2−2y2, so we have
d
dx (2y′)−(−4y) = 0
which simplifies to d
dx (2y′)+4y= 0
Step 2: Solve the differential equation. Rewriting the equation in a more
familiar form, we have
2y′′ + 4y= 0
The characteristic equation is 2r2+ 4 = 0, which gives us r=±i.
Therefore, the general solution is
y(x) = c1cos(2x) + c2sin(2x)
Step 3: Apply boundary conditions.
From the first boundary condition y(0) = 1, we have c1= 1. From the
second boundary condition y(1) = 2, we have c1cos(2) + c2sin(2) = 2.
Substitute c1= 1 into the second boundary condition to get
cos(2) + c2sin(2) = 2
Solving for c2, we get
c2=2−cos(2)
sin(2)
Step 4: Final solution.
Therefore, the extremal of the functional is
y(x) = cos(2x) + 2−cos(2)
sin(2) sin(2x)
Question 16
Question
Find the extremals of the functional
J[y] = Z1
0
(y2+y′2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
15
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
the functional J[y] = Rb
aL(x, y, y′)dx is given by
d
dx
∂L
∂y′−∂L
∂y = 0.
Step 2: Given the functional J[y] = R1
0(y2+y′2)dx, we have L(x, y, y′) =
y2+y′2. The Euler-Lagrange equation becomes
d
dx
∂(y2+y′2)
∂y′−∂(y2+y′2)
∂y = 0.
Step 3: Compute the partial derivatives. We have
∂(y2+y′2)
∂y′= 2y′,
∂(y2+y′2)
∂y = 2y.
Substituting these into the Euler-Lagrange equation gives
d
dx(2y′)−2y= 0.
Step 4: Simplify the Euler-Lagrange equation. This simplifies to
2y′′ −2y= 0.
Step 5: Solve the differential equation. The general solution to 2y′′ −2y= 0
is y(x) = c1cos√2x+c2sin√2x.
Step 6: Apply the boundary conditions. Using y(0) = 0, we find c1= 0.
Using y(1) = 1, we find c2=1
sin(√2).
Step 7: The extremal for the functional is
y(x) = sin√2x
sin√2.
Question 17
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
16
Solution
Step 1: Compute the Euler-Lagrange equation. The Euler-Lagrange equation
for a functional of the form J[y] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
In this case, F(x, y, y′) = y′2−y2, so
∂F
∂y =−2yand ∂F
∂y′= 2y′
d
dx ∂F
∂y′=d
dx(2y′)=2y′′
Therefore, the Euler-Lagrange equation is
2y′′ + 2y= 0
Step 2: Solve the differential equation. The general solution to the differen-
tial equation 2y′′ + 2y= 0 is
y(x) = c1cos(x) + c2sin(x)
Step 3: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we have
y(0) = c1= 0
y(1) = c2sin(1) = 1
c2=1
sin(1)
Therefore, the extremal that minimizes the functional is
y(x) = 1
sin(1) sin(x)
Question 18
Question
Find the extremals for the functional
J[y] = Z1
0
(y′2+y′′)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
17
Solution
Step 1: Compute the Euler-Lagrange equation. Given the functional
J[y] = Z1
0
(y′2+y′′)dx
We have the Euler-Lagrange equation
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2+y′′.
Step 2: Find the partial derivatives and substitute into the Euler-Lagrange
equation. We have ∂f
∂y′=∂
∂y′(y′2+y′′)=2y′
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
∂f
∂y =∂
∂y (y′2+y′′)=0
Substitute these derivatives into the Euler-Lagrange equation:
2y′′ −0=0
y′′ = 0
Step 3: Solve the differential equation y′′ = 0. The general solution to y′′ = 0
is y=Ax +B, where Aand Bare constants.
Step 4: Apply the boundary conditions. Since y(0) = 0 and y(1) = 1, we
have the following system of equations:
B= 0
A+B= 1
From the first equation, we have B= 0, and substituting into the second
equation gives A= 1. Therefore, the extremal that minimizes the functional is
y=x.
Question 19
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
18
Solution
Step 1: Compute the Euler-Lagrange equation. This equation is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y.
Step 2: Compute the partial derivatives:
∂f
∂y′=∂(y′2−y)
∂y′= 2y′
∂f
∂y =∂(y′2−y)
∂y =−1
Step 3: Plug the derivatives into the Euler-Lagrange equation:
d
dx (2y′) + 1 = 0
Step 4: Simplify the equation:
2y′′ + 1 = 0
Step 5: Solve the differential equation:
2y′′ =−1
y′′ =−1
2
Step 6: Integrate y′′ twice to find y:
y′=−x
2+A
y=−x2
4+Ax +B
Step 7: Apply the boundary conditions to solve for Aand B:
y(0) = 0 ⇒B= 0
y(1) = 1 ⇒ −1
4+A= 1 ⇒A=5
4
Therefore, the extremal for the functional J[y] that satisfies the boundary
conditions is
y=−x2
4+5
4x
19
Question 20
Question
Let J[y] = R1
0y(x)−y′(x)2dx where y(0) = 0 and y(1) = 1. Find the function
y(x) that minimizes J[y].
Solution
Step 1: Write down the Euler-Lagrange equation for the given functional J[y]:
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y−y′(x)2.
Step 2: Compute the partial derivatives:
∂f
∂y = 1 and ∂f
∂y′=−2y′
Step 3: Apply the Euler-Lagrange equation:
d
dx(−2y′)−1=0
Simplify the equation to get:
2y′′ −1=0
Step 4: Solve the differential equation 2y′′ −1 = 0 subject to the boundary
conditions y(0) = 0 and y(1) = 1.
Step 5: Integrate twice to find the general solution:
y′′ =1
2=⇒y′=1
2x+C1
y=1
4x2+C1x+C2
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1 to find C1
and C2:
0 = C2and 1 = 1
4+C1
Thus, C1=3
4.
Step 7: Substitute C1back into yto get the specific function that minimizes
J[y]:
y=1
4x2+3
4x
Therefore, the function y(x) = 1
4x2+3
4xminimizes the functional J[y].
20
Question 21
Question
Find the extremals for the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
To find the extremals for the functional J[y], we will apply the Euler-Lagrange
equation.
Step 1: Formulating the Euler-Lagrange equation Let F(x, y, y′) =
y′2−y2. The Euler-Lagrange equation is given by
d
dx ∂F
∂y′−∂F
∂y = 0
Step 2: Computing the partial derivatives Compute the partial deriva-
tives of F:∂F
∂y′= 2y′,∂F
∂y =−2y
Step 3: Applying the Euler-Lagrange equation Substitute the partial
derivatives into the Euler-Lagrange equation:
d
dx(2y′)−(−2y)=0
d
dx(2y′)+2y= 0
Step 4: Solving the differential equation Integrate the above equation:
2y′+ 2y=C
y′+y=C1
where C1is an arbitrary constant.
Step 5: Determining the extremals Applying the boundary conditions
y(0) = 0 and y(1) = 1:
y′(0) + y(0) = C1= 0
y′(1) + y(1) = C1= 1
Thus, the extremal that satisfies these boundary conditions is y(x) = e−x.
Therefore, the extremal for the functional J[y] is y(x) = e−x.
21
Question 22
Question
Consider the functional J(y) = R1
0(y′2−y2)dx where the function ysatisfies the
boundary conditions y(0) = 0 and y(1) = 1. Find the function ythat minimizes
J(y).
Solution
Step 1: Calculate the Euler-Lagrange equation by differentiating the integrand
with respect to yand y′.
d
dx ∂f
∂y′−∂f
∂y = 0
In this case, the Lagrangian f=y′2−y2, so ∂f
∂y =−2yand ∂f
∂y′= 2y′. Then,
d
dx(2y′)−(−2y)=0
2y′′ + 2y= 0
y′′ +y= 0
Step 2: Solve the differential equation y′′ +y= 0 subject to the boundary
conditions y(0) = 0 and y(1) = 1.
The general solution to the differential equation is of the form y(x) =
Asin x+Bcos x. Applying the boundary conditions: y(0) = 0 ⇒B= 0
y(1) = 1 ⇒Asin 1 = 1 ⇒A=1
sin 1
Therefore, the function ythat minimizes the functional J(y) is y(x) = sin x
sin 1 .
Question 23
Question
Let J[y] = R1
0(y′2−y2)dx be a functional defined on the set of twice continuously
differentiable functions y(x) satisfying y(0) = 0 and y(1) = 1. Find the function
y(x) that minimizes J[y].
Solution
Step 1: Let L(y, y′, x) = y′2−y2be the integrand of the functional J[y].
Step 2: The Euler-Lagrange equation for minimizing J[y] is given by
d
dx ∂L
∂y′−∂L
∂y = 0
22
Step 3: Compute the partial derivatives:
∂L
∂y′= 2y′
∂L
∂y =−2y
Step 4: Apply the Euler-Lagrange equation:
d
dx (2y′)−(−2y) = 0
2y′′ + 2y= 0
Step 5: Solve the differential equation: The general solution to y′′ +y= 0
is of the form y(x) = Asin(x) + Bcos(x).
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1:
y(0) = B= 0
y(1) = Asin(1) = 1
A=1
sin(1)
Step 7: Therefore, the function y(x) = sin(x)
sin(1) minimizes the functional J[y]
subject to the given boundary conditions.
Question 24
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2+yy′+y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
Step 1: Calculate the Euler-Lagrange equation by differentiating the integrand
with respect to yand y′.
∂f
∂y −d
dx ∂f
∂y′= 0
where f=y′2+yy′+y2.
23
Step 2: Compute ∂f
∂y and ∂f
∂y′.
∂f
∂y = 2y+y′
∂f
∂y′= 2y′+y
Step 3: Differentiate ∂f
∂y′with respect to x.
d
dx ∂f
∂y′=d
dx(2y′+y)=2y′′ +y′
Step 4: Set up the Euler-Lagrange equation and simplify.
2y+y′−(2y′′ +y′) = 0
2y+y′−2y′′ −y′= 0
2y−2y′′ = 0
y−y′′ = 0
Step 5: Solve the differential equation y−y′′ = 0 with the boundary condi-
tions y(0) = 0 and y(1) = 1. The general solution to the differential equation is
y(x) = c1ex+c2e−x. Using the boundary conditions:
y(0) = c1+c2= 0
y(1) = c1e+c2e−1= 1
Step 6: Solve the system of equations to find the constants c1and c2.
c1=e
e−1, c2=−1
e−1
Thus, the extremal for the given functional is y(x) = e
e−1ex−1
e−1e−x.
Question 25
Question
Find the extremals of the functional
J[y] = Z1
0
(2y2+y′2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
24
Solution
Step 1: Define the Euler-Lagrange equation.
The Euler-Lagrange equation for a functional J[y] = Rb
aL(x, y, y′)dx is given
by: d
dx
∂L
∂y′=∂L
∂y
Step 2: Find the partial derivatives of L(x, y, y′).
In this case, L(x, y, y′)=2y2+y′2, so:
∂L
∂y = 4yand ∂L
∂y′= 2y′
Step 3: Calculate the derivative d
dx
∂L
∂y′.
Taking the derivative of ∂L
∂y′= 2y′with respect to x:
d
dx
∂L
∂y′=d
dx(2y′)=2y′′
Step 4: Apply the Euler-Lagrange equation.
By applying the Euler-Lagrange equation, we have:
2y′′ = 4y
This simplifies to the second-order differential equation y′′ −2y= 0.
Step 5: Solve the differential equation y′′ −2y= 0.
The characteristic equation for this differential equation is r2−2 = 0, which has
solutions r=±√2. Therefore, the general solution to the differential equation
is:
y(x) = c1e√2x+c2e−√2x
Step 6: Apply the boundary conditions y(0) = 0 and y(1) = 1.
Using the boundary conditions y(0) = 0 and y(1) = 1, we find:
0 = c1+c2
1 = c1e√2+c2e−√2
Solving this system of equations, we get c1=e−√2
e−√2−e√2and c2=−e√2
e−√2−e√2.
Step 7: Determine the extremal function.
Substitute the values of c1and c2back into the general solution to obtain the
extremal function y(x).
Therefore, the extremals of the functional J[y] = R1
0(2y2+y′2)dx subject
to the boundary conditions y(0) = 0 and y(1) = 1 are the solutions to the
differential equation y′′ −2y= 0 with the extremal function derived from the
boundary conditions.
25
Question 26
Question
Find the extremals for the functional
J[y] = Z1
0
(2y−y′2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
Solution
To find the extremals for the given functional J[y], we will use the Euler-
Lagrange equation:
d
dx ∂f
∂y′−∂f
∂y = 0
where f= 2y−y′2.
Step 1: Find ∂f ∂y and ∂f
∂y′
∂f
∂y = 2
∂f
∂y′=−2y′
Step 2: Apply the Euler-Lagrange equation
d
dx(−2y′)−2=0
d
dx(−2y′)=2
−2y′′ = 2
Step 3: Solve the ODE Integrate both sides to solve for y(x):
−2y′= 2x+C1
y′=−x−C1/2
y=−1
2x2−C1x+C2
Step 4: Apply the boundary conditions Using the boundary conditions
y(0) = 0 and y(1) = 1:
y(0) = C2= 0
y(1) = −1
2+C1= 1
26
C1=3
2
Step 5: Final Solution The extremal for the given functional subject to
the boundary conditions is:
y(x) = −1
2x2+3
2x
Question 27
Question
Find the extremals of the functional
J[y] = Z1
0
(y′(x))2+y(x)2dx
subject to the boundary conditions y(0) = 0 and y(1) = 3.
Solution
Step 1: Define the Euler-Lagrange equation The Euler-Lagrange equation for
the given functional J[y] is
d
dx ∂f
∂y′−∂f
∂y = 0,
where f(y, y′, x)=(y′(x))2+y(x)2.
Step 2: Calculate the partial derivatives Let’s first find the partial derivatives
of fwith respect to yand y′:
∂f
∂y = 2y(x),∂f
∂y′= 2y′(x).
Step 3: Apply the Euler-Lagrange equation Substitute the partial derivatives
into the Euler-Lagrange equation:
d
dx(2y′(x)) −2y(x)=0
2y′′(x)−2y(x) = 0.
Step 4: Solve the differential equation The general solution to the differential
equation 2y′′(x)−2y(x) = 0 is
y(x) = c1cos(x) + c2sin(x).
Step 5: Apply the boundary conditions Using the boundary conditions
y(0) = 0 and y(1) = 3, we find:
(y(0) = c1= 0
y(1) = c1cos(1) + c2sin(1) = 3
27
Step 6: Solve for the constants From c1= 0, we have c2sin(1) = 3, so
c2=3
sin(1) . Thus, the extremal of the functional is
y(x) = 3
sin(1) sin(x).
Question 28
Question
Find the extremal for the functional J[y] = R1
0y′(x)2+y(x)2dx subject to the
constraint R1
0y(x)dx = 1.
Solution
Step 1: Define the Lagrangian. Let L=y′(x)2+y(x)2+λ(y(x)−1), where λ
is a Lagrange multiplier.
Step 2: Find the Euler-Lagrange equation. The Euler-Lagrange equation
for this problem is given by
d
dx ∂L
∂y′−∂L
∂y = 0.
This gives us
d
dx(2y′(x)) −2y(x) + λ= 0 and 2y(x)+2λ= 0.
Step 3: Solve for y(x) using the boundary condition. From the second
equation in Step 2, we have y(x) = −λ. Substituting this back into the first
equation and integrating, we find y(x) = sin 2πx.
Step 4: Use the constraint to find λ. Apply the constraint R1
0y(x)dx = 1:
Z1
0
sin 2πx dx =1
2πcos 2πx
1
0=1
π= 1.
Thus, λ=−1.
Step 5: Final extremal. Therefore, the extremal for the given functional
subject to the constraint is y(x) = −sin 2πx.
Question 29
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
28
Solution
Step 1: Define the Lagrangian function. We define the Lagrangian function as
L(x, y, y′) = y′2−y2
Step 2: Set up the Euler-Lagrange equation. The Euler-Lagrange equation
for the functional J[y] is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Applying this to our Lagrangian L(x, y, y′) = y′2−y2, we get
d
dx(2y′)+2y= 0
Step 3: Solve the Euler-Lagrange equation. Solving the above differential
equation gives us
2y′′ + 2y= 0
Step 4: Find the general solution to the differential equation. The general
solution to the differential equation 2y′′ + 2y= 0 is
y(x) = c1cos(x) + c2sin(x)
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find
c1= 0
c2sin(1) = 1
c2=1
sin(1)
Step 6: Write down the extremal. Therefore, the extremal of the functional
J[y] that satisfies the given boundary conditions is
y(x) = 1
sin(1) sin(x)
Question 30
Question
Find the extremals for the functional
J[y] = Z2
1
(y′2−2y)dx
subject to the boundary conditions y(1) = 0 and y(2) = 1.
29
Solution
Step 1: Compute the Euler-Lagrange equation. The Euler-Lagrange equation
for the functional J[y] is given by
d
dx ∂f
∂y′−∂f
∂y = 0
where f(y, y′, x) = y′2−2y.
Step 2: Find the partial derivatives. We have
∂f
∂y =−2,and ∂f
∂y′= 2y′
and d
dx ∂f
∂y′=d
dx(2y′)=2y′′
Step 3: Form the Euler-Lagrange equation. Substitute the partial derivatives
into the Euler-Lagrange equation to get
2y′′ + 2 = 0
Step 4: Solve the differential equation. The general solution to the differen-
tial equation is
y(x) = c1+c2x−x2
where c1and c2are constants to be determined.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(1) = 0 and y(2) = 1, we have
c1+c2−1=0
c1+ 2c2−4=1
Solving these equations simultaneously, we find c1=−2 and c2= 3.
Therefore, the extremal for the functional J[y] is
y(x) = −2+3x−x2
Question 31
Question
Find the extremal of the functional
J[y] = Z1
0
(y′2−y2)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
30
Solution
Step 1: We define the Euler-Lagrange equation as
d
dx ∂f
∂y′−∂f
∂y = 0
where f=y′2−y2.
Step 2: Calculate the partial derivatives:
∂f
∂y′= 2y′and ∂f
∂y =−2y
Step 3: Compute the derivative with respect to xin the Euler-Lagrange
equation:
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
Step 4: Substitute the derivatives back into the Euler-Lagrange equation:
2y′′ −(−2y)=0
2y′′ + 2y= 0
Step 5: Solve the differential equation 2y′′ + 2y= 0 for y(x) with the bound-
ary conditions y(0) = 0 and y(1) = 1.
Step 6: The general solution to the differential equation is y(x) = Asin(x) +
Bcos(x).
Step 7: Use the boundary conditions to find Aand B:
y(0) = 0 =⇒B= 0
y(1) = 1 =⇒Asin(1) = 1 =⇒A=1
sin(1)
Step 8: Therefore, the extremal of the functional J[y] subject to the given
boundary conditions is
y(x) = sin(x)
sin(1)
Question 32
Question
Find the extremals of the functional
J[y] = Z1
0
(y′2−y)dx
subject to the boundary conditions y(0) = 0 and y(1) = 1.
31
Solution
Step 1: Define the Euler-Lagrange equation. The Euler-Lagrange equation for
the functional J[y] = Rb
aL(x, y, y′)dx is given by
d
dx ∂L
∂y′−∂L
∂y = 0
Step 2: Calculate the partial derivatives. In this case, we have L(x, y, y′) =
y′2−y. Therefore, the partial derivatives are:
∂L
∂y =−1
∂L
∂y′= 2y′
d
dx ∂L
∂y′= 2y′′
Step 3: Set up the Euler-Lagrange equation. Substitute the partial deriva-
tives into the Euler-Lagrange equation to get:
d
dx(2y′) + 1 = 0
Step 4: Solve the differential equation. The differential equation simplifies
to 2y′′ + 1 = 0. Solving this gives y(x) = 1
2x2+Ax +B.
Step 5: Apply the boundary conditions. Using the boundary conditions
y(0) = 0 and y(1) = 1, we find B= 0 and A=1
2. Therefore, the extremal is
y(x) = 1
2x2+1
2x.
Step 6: Check the second variation. To confirm that this extremal is a
minimum, we need to check the second variation. This involves computing the
second variation of the functional and verifying the sign of δ2J.
Question 33
Question
Evaluate the functional
J[y] = Z1
02y2+y′2dx
subject to the boundary conditions y(0) = 0 and y(1) = 0.
32
Solution
Step 1: Compute the Euler-Lagrange equation by differentiating the integrand
with respect to yand y′.
∂f
∂y −d
dx ∂f
∂y′= 0
where f= 2y2+y′2.
Step 2: Compute ∂f
∂y .
∂f
∂y = 4y
Step 3: Compute ∂f
∂y′.
∂f
∂y′= 2y′
Step 4: Compute d
dx ∂f
∂y′.
d
dx ∂f
∂y′=d
dx(2y′)=2y′′
Step 5: Substitute the results back into the Euler-Lagrange equation.
4y−2y′′ = 0
Step 6: Rearrange the equation to get a differential equation.
2y′′ −4y= 0
Step 7: Solve the differential equation by assuming a solution of the form
y(x) = erx.
2r2erx −4erx = 0
Step 8: Simplify the equation and solve for r.
2r2−4 = 0 =⇒r=±√2
Step 9: The general solution is of the form y(x) = c1e√2x+c2e−√2x.
Step 10: Apply the boundary conditions y(0) = 0 and y(1) = 0 to find the
specific solution.
c1+c2= 0
c1e√2+c2e−√2= 0
Step 11: Solve the system of equations to find c1and c2.
c1=e−√2
e√2−e−√2=−1
2√2
c2=e√2
e√2−e−√2=1
2√2
Step 12: Therefore, the function that minimizes the functional J[y] is y(x) =
−1
2√2e√2x+1
2√2e−√2x.
33
Question 34
Question
Consider the functional J[y] = R1
0(y′(x))2+y(x)dx subject to the boundary
conditions y(0) = 1 and y(1) = 2. Find the function y(x) that minimizes J[y].
Solution
Step 1: Define the Lagrangian. The Lagrangian for the given functional J[y]
with the constraint of fixed boundary conditions can be defined as:
L(y, y′;λ)=(y′(x))2+y(x) + λ1(y(0) −1) + λ2(y(1) −2)
Step 2: Find the Euler-Lagrange equation. The Euler-Lagrange equation for
minimizing the functional J[y] is given by:
d
dx ∂L
∂y′−∂L
∂y = 0
Step 3: Solve the Euler-Lagrange equation. Applying the Euler-Lagrange
equation to the Lagrangian L, we get:
d
dx(2y′)−1=0
2y′′ −1=0
Step 4: Solve the differential equation. Solving the differential equation
2y′′ −1 = 0 with the boundary conditions y(0) = 1 and y(1) = 2, we find:
y(x) = 1
2x2+3
2x+ 1
Step 5: Verify the solution. To verify that this function minimizes the
functional J[y], we need to check if it satisfies the boundary conditions and
compare its value with other candidate functions.
Question 35
Question
Find the extremals for the functional
J[y] = Z2
1
(y′2−2y)dx
subject to the boundary conditions y(1) = 1 and y(2) = 2.
34
Solution
Step 1: Compute the Euler-Lagrange equation. The Euler-Lagrange equation
for the functional J[y] = Rb
aF(x, y, y′)dx is given by
d
dx ∂F
∂y′−∂F
∂y = 0
In this case, F(x, y, y′) = y′2−2y. Thus,
∂F
∂y =−2 and ∂F
∂y′= 2y′
Therefore, the Euler-Lagrange equation becomes
d
dx(2y′) + 2 = 0
2y′′ + 2 = 0
Step 2: Solve the differential equation. Solving the differential equation
2y′′ + 2 = 0 gives us
y′′ =−1
Integrating once gives
y′=−x+c1
Integrating again gives
y=−1
2x2+c1x+c2
Step 3: Apply the boundary conditions. Using the first boundary condition
y(1) = 1,
1 = −1
2(1)2+c1(1) + c2
1 = −1
2+c1+c2
Using the second boundary condition y(2) = 2,
2 = −1
2(2)2+c1(2) + c2
2 = −2+2c1+c2
Solving the system of equations, we find c1=3
2and c2= 1.
Therefore, the extremal function is
y=−1
2x2+3
2x+ 1
35
